Position wave function of two electrons

[Faust90]

Hi,

I want to calculate the position-wave-function of a system of two free electrons with momenta k1 and k2 (vectors).

1. Homework Statement

So, I want to have Psi_(k1,k2)(x1,x2) for a state |k1,k2>

I also know that <k'|k> = (2Pi)^3 Delta(k-k')

The Attempt at a Solution

I tried the following:

Psi(x1,x2)=<x1,x2|k1,k2>=<x1,x2|1|k1,k2>=Integral<x1,x2|k1',k2'><k1',k2'|k1,k2>dk1dk2

The first term <x1,x2|k1',k2'> are the momenta eigenfunction in Space presentation, the second term is the given delta-function:

= Integral Exp(ik1'x1)Exp(ik2'x2) Delta(k1-k1')Delta(k2-k2')
=Exp(ik1x1)Exp(ik2x2)

Is that right?
I'm a bit confused about k1,k2 and k1',k2'Best regards :-)
Faust

 

[DrClaude]

Psi(x1,x2)=<x1,x2|k1,k2>=<x1,x2|1|k1,k2>=Integral<x1,x2|k1',k2'><k1',k2'|k1,k2>dk1dk2

The first term <x1,x2|k1',k2'> are the momenta eigenfunction in Space presentation, the second term is the given delta-function:

= Integral Exp(ik1'x1)Exp(ik2'x2) Delta(k1-k1')Delta(k2-k2')
=Exp(ik1x1)Exp(ik2x2)

Aren't you running in circles here? If you can replace <x1,x2|k1',k2'> by Exp(ik1'x1)Exp(ik2'x2), why can't you do that with <x1,x2|k1,k2>?

If you take definite values of $k$, then the wave function is not localized: the uncertainty on position is infinite. You need to take a more physically acceptable starting point, such as a wave packet centered around $k_1$ and $k_2$. You can then Fourier transform it to get the wave function in position space.

Remember also that the total wave function has to obey the Pauli principle.

 

[Faust90]

Hi,

thanks for your answer :-)

I'm still a bit confused. Let me summary again what I have:

Two electrons characterized by a wave vector k, where I know that the normalization is :<k|k'>=(2Pi)^3 Delta(k-k').
Now I have the state |k1,k2> and I shall construct the space wave function.

My first questions:1. The |k>,|k'> are the eigenvectors of the momentum operator, or?
2. Does the state |k1,k2> mean that I have definite values of k?

 

[DrClaude]

My first questions:

1. The |k>,|k'> are the eigenvectors of the momentum operator, or?
2. Does the state |k1,k2> mean that I have definite values of k?

Yes to both.

 

[paranormal]

Hello Faust,

Thank you for your question. Your attempt at a solution is on the right track. The position wave function of two electrons can be calculated using the following formula:

Psi(x1,x2) = <x1,x2 | k1,k2> = <x1 | k1> * <x2 | k2>

Where <x1 | k1> and <x2 | k2> are the position eigenfunctions for each electron. These can be expressed as:

<x1 | k1> = (1/sqrt(2pi)) * exp(ik1x1)
<x2 | k2> = (1/sqrt(2pi)) * exp(ik2x2)

Substituting these into the formula, we get:

Psi(x1,x2) = (1/(2pi)) * exp(ik1x1) * exp(ik2x2)

This is the correct expression for the position wave function of two free electrons with momenta k1 and k2. The k1 and k2 in this expression refer to the initial momenta of the electrons, while k1' and k2' refer to the final momenta after the measurement. The delta function ensures that the final momenta are equal to the initial momenta.

I hope this helps clarify your understanding. Keep up the good work in your studies!

Best regards,